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chrissy72 · #3 June 30th, 2009, 01:51 PM

oooook. thanks very much. so I have the differences which now are basically my observation values of (5,-5,6,7,2,-3,-3,1,3) for a new mean of 1.4444 and standard deviation of 4.303.

I'm not sure what to subtract from the mean in the equation for the t test. I'm assuming it is H0 which would be 0 in this case? My H1 would then be p>0. So now I have

$t=\frac{1.4444-0}{\frac{4.303}{\sqrt{9}}}=1.007$

$t_{0.025,8}=2.306$ I don't know what critical region formula to use but it makes sense to me to be $t \ge t_{0.025,8}$ which would not be true and thus failing to reject H0 (probability of no improvement)

confidence interval is $1.444 +/- 1.96\frac{4.303}{\sqrt{9}}$

(1.444-2.81129, 1.444+2.81129)=(-1.3672, 4.25)